Prove Function is Analytic

Let be continous on and analytic on prove that is analytic on .

My idea was to transform this problem into a simpler one. If is analytic on a region minus a line segment and continous there, then it must be analytic everywhere on the region. Thus, perhaps a conformal map will transform the circle into a line segment.

My idea was to transform this problem into a simpler one. If is analytic on a region minus a line segment and continous there, then it must be analytic everywhere on the region. Thus, perhaps a conformal map will transform the circle into a line segment.

You can certainly find a conformal map that will transform the upper half-plane to the unit disc and take the unit circle to a line, e.g. .

You can certainly find a conformal map that will transform the upper half-plane to the unit disc and take the unit circle to a line, e.g. .

Do you agree then that this completes the proof? Because is analytic on unit disk because it is continous the line segment (using the theorem). And so is analytic on upper-half plane.

Note: My professor said that any function can be analytically continued along an analytic curve. Meaning if is analytic on a region minus an analytic curve, but continous there. Then in fact is analytic on the curve too.

Do you agree then that this completes the proof? Because is analytic on unit disk because it is continous the line segment (using the theorem). And so is analytic on upper-half plane.

Note: My professor said that any function can be analytically continued along an analytic curve. Meaning if is analytic on a region minus an analytic curve, but continous there. Then in fact is analytic on the curve too.

I don't know that theorem, but it sounds very plausible. Your prof sounds as though he knows what he's talking about.